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Difference between revisions of "2007 AMC 10B Problems/Problem 8"
 
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CHIKEN NUGGIEs
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On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form <math>bbcac,</math> where <math>0 \le a < b < c \le 9,</math> and <math>b</math> was the average of <math>a</math> and <math>c.</math> How many different five-digit numbers satisfy all these properties?
  
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==Solution 2==
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<math>\textbf{(A) } 12 \qquad\textbf{(B) } 16 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 25</math>
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==Solution==
  
 
Case <math>1</math>: The numbers are separated by <math>1</math>.
 
Case <math>1</math>: The numbers are separated by <math>1</math>.
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It's pretty clear that there's a pattern: <math>8</math> sets, <math>6</math> sets, <math>4</math> sets. The amount of sets per case decreases by <math>2</math>, so it's obvious Case <math>4</math> has <math>2</math> sets. The total amount of possible five-digit numbers is <math>8+6+4+2=\boxed{\textbf{(D)}\ 20}</math>.
 
It's pretty clear that there's a pattern: <math>8</math> sets, <math>6</math> sets, <math>4</math> sets. The amount of sets per case decreases by <math>2</math>, so it's obvious Case <math>4</math> has <math>2</math> sets. The total amount of possible five-digit numbers is <math>8+6+4+2=\boxed{\textbf{(D)}\ 20}</math>.
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== See Also ==
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{{AMC10 box|year=2007|ab=B|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 19:51, 19 April 2024

On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form $bbcac,$ where $0 \le a < b < c \le 9,$ and $b$ was the average of $a$ and $c.$ How many different five-digit numbers satisfy all these properties?

$\textbf{(A) } 12 \qquad\textbf{(B) } 16 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 25$

Solution

Case $1$: The numbers are separated by $1$.

We this case with $a=0, b=1,$ and $c=2$. Following this logic, the last set we can get is $a=7, b=8,$ and $c=9$. We have $8$ sets of numbers in this case.


Case $2$: The numbers are separated by $2$.

This case starts with $a=0, b=2,$ and $c=2$. It ends with $a=5, b=7,$ and $c=9$. There are $6$ sets of numbers in this case.


Case $3$: The numbers start with $a=0, b=3,$ and $c=6$. It ends with $a=3, b=6,$ and $c=9$. This case has $4$ sets of numbers.

It's pretty clear that there's a pattern: $8$ sets, $6$ sets, $4$ sets. The amount of sets per case decreases by $2$, so it's obvious Case $4$ has $2$ sets. The total amount of possible five-digit numbers is $8+6+4+2=\boxed{\textbf{(D)}\ 20}$.

See Also

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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